Solution of a repeating polynomial I have a polynomial series
$$ P_n(x) = x(1-P_{n-1}(x)) $$
With initial value $ P_1 = x $, how can I solve this as a sum?
$$ \Sigma_{n=1}^\infty nP_n(x)$$
 A: Let $ n $ be a positive integer greater than $ 1 $, we have for all $ k\in\mathbb{N}^{*} $ less than $ n $ : \begin{aligned} P_{n-k}\left(x\right)&=x\left(1-P_{n-k-1}\left(x\right)\right)\\ &\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(-x\right)^{k}P_{n-k}\left(x\right)&=x\left(-x\right)^{k}+\left(-x\right)^{k+1}P_{n-k-1}\left(x\right)\\ \iff \sum_{k=0}^{n-2}{\left(\left(-x\right)^{k}P_{n-k}\left(x\right)-\left(-x\right)^{k+1}P_{n-k-1}\left(x\right)\right)}&=x\sum_{k=0}^{n-2}{\left(-x\right)^{k}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{n}\left(x\right)-\left(-x\right)^{n-1}P_{1}\left(x\right)&=\frac{x+\left(-x\right)^{n}}{1+x} \end{aligned}
Thus : $$ \left(\forall n\in\mathbb{N}^{*}\right),\ P_{n}\left(x\right)=\frac{x+\left(-x\right)^{n+1}}{1+x} $$
Hence your series doesn't converge for all reals $ x $ except when $ x $ is $ 0 \cdot $
A: Let's put first
$$
\eqalign{
  & P_{\,n} (x) = x\left( {1 - P_{\,n - 1} (x)} \right)\quad \left| {\;P_{\,1} (x) = x} \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad Q_{\,n} (x) = x\left( {1 - Q_{\,n - 1} (x)} \right)\quad \left| \matrix{
  \;P_{\,n + 1} (x) = Q_{\,n} (x) \hfill \cr 
  \;Q_{\,0} (x) = x \hfill \cr}  \right. \cr} 
$$
Then let's take the z-Transform of both sides
$$
\eqalign{
  & F(x,z) = \sum\limits_{0\, \le \,n} {Q_{\,n} (x)z^{\,n} }  = \sum\limits_{0\, \le \,n} {x\left( {1 - Q_{\,n - 1} (x)} \right)z^{\,n} }  =   \cr 
  &  = x\sum\limits_{0\, \le \,n} {z^{\,n} }  - xz\sum\limits_{0\, \le \,n} {Q_{\,n - 1} (x)z^{\,n - 1} }  =   \cr 
  &  = {x \over {1 - z}} - xzF(x,z)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad F(x,z) = {x \over {\left( {1 - z} \right)\left( {1 + xz} \right)}}
 = {{x^{\,2} } \over {\left( {1 + x} \right)\left( {1 + xz} \right)}} + {x \over {\left( {1 + x} \right)\left( {1 - z} \right)}} \cr} 
$$
That means that
$$
Q_{\,n} (x) = {x \over {1 + x}}\left( {x\left( { - 1} \right)^{\,n} x^{\,n}  + 1} \right)
 = \left( { - 1} \right)^{\,n} {x \over {1 + x}}\left( {x^{\,n + 1}  - \left( { - 1} \right)^{\,n + 1} } \right)
$$
And finally
$$
\eqalign{
  & Q_{\,2n} (x) = x{{x^{\,2n + 1}  + 1} \over {1 + x}} = x{{1 - \left( { - x} \right)^{\,2n + 1} } \over {1 - \left( { - x} \right)}} =   \cr 
  &  = x\sum\limits_{k = 0}^{2n} {\left( { - x} \right)^k }  = \sum\limits_{k = 0}^{2n} {\left( { - 1} \right)^k x^{k + 1} }   \cr 
  & Q_{\,2n + 1} (x) =  - x{{x^{\,2n + 2}  - 1} \over {1 + x}} =  - x{{\left( {x^{\,n + 1}  + 1} \right)\left( {x^{\,n + 1}  - 1} \right)} \over {1 + x}} =   \cr 
  &  =  - x\left( {x^{\,n + 1}  - 1} \right){{\left( {1 - \left( { - x} \right)^{\,n + 1} } \right)} \over {1 - \left( { - x} \right)}}
 = \left( {1 - x^{\,n + 1} } \right)\sum\limits_{k = 0}^n {\left( { - 1} \right)^k x^{k + 1} }  \cr} 
$$
